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anyanavicka [17]
3 years ago
14

Differentiate between gravity andgravitation​

Physics
2 answers:
wariber [46]3 years ago
8 0

Answer:

Gravitation is the attractive force existing between any two objects that have mass. The force of gravitation pulls objects together. Gravity is the gravitational force that occurs between the earth and other bodies. Gravity is the force acting to pull objects toward the earth.

Explanation:

ZanzabumX [31]3 years ago
6 0
Gravitation is the attractive force existing between any two objects that have mass. The force of gravitation pulls objects together. Gravity is the gravitational force that occurs between the earth and other bodies. Gravity is the force acting to pull objects toward the earth.
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Which of the following advancements could science and technology reasonably contribute to society?
Alexxx [7]

Answer:            

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Explanation:

7 0
3 years ago
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A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi
stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

v=\sqrt{\frac{TL}{M}} (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

v= \lambda f (2)

We can equate expression (1) and (2):

\sqrt{\frac{TL}{M}}=\lambda f

Solving for T

T= \frac{M(\lambda f)^2}{L} (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic N_{harmonic} is:

\lambda_{harmonic}=\frac{2l}{N_{harmonic}}

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m

We can now find T on (3) using all the values we have:

T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}

T=75.22 N

3 0
3 years ago
How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0
3 years ago
Find the angle for the third-order maximum for 591 nm wavelength light falling on a diffraction grating having 1460 lines per ce
Marat540 [252]

Answer:

15.32°

Explanation:

We have given the wavelength \lambda =591nm=591\times 10^{-9}m

Diffraction grating is 1460 lines per cm

So  d=\frac{10^{-2}}{1460}=6.71\times 10^{-6}m (as 1 m=100 cm )

For maximum diffraction

dsin\Theta =m\lambda here m is order of diffraction

So 6.71\times 10^{-6}sin\Theta =3\times 591\times 10^{-9}

sin\Theta =0.264

\Theta =15.32^{\circ}

6 0
2 years ago
Question 29 of 43
liraira [26]
I agree with the first responses
3 0
2 years ago
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